Assume further that \(d^2=o(n)\) . Letting \(b_n:=n/d\) , we see that \(b_n\rightarrow \infty \)  and \(nb_n^{-2}\rightarrow 0\) . By classical weak laws of large numbers (see Theorem 2 in [1]} with a discussion above it and page 317 in [2]}), \((S_n-n)/b_n=o_{\mathbb {P}}(1)\)  if and only if
\(n{\mathbb {P}}( |Z-1|>b_n )+nb_n^{-2} {\mathbb {E}}(Z-1)^2 \mathbf {1}( |Z-1|\leqslant b_n )+nb_n^{-1} |{\mathbb {E}}(Z-1) \mathbf {1}(|Z-1|\leqslant b_n )|\rightarrow 0 .\) 
